Pressure in open tube

In the L shape tube shown in the image shouldn't the pressure at all points A, B,C,D be same as atmospheric pressure? I have to find relation between A, B, C and D

• Yes, the pressures at A, B, C, and D are atmospheric. So,....? Commented Aug 23, 2021 at 12:01

The "$$\omega$$" in the drawing suggests that the tube is rotating. In that case there will be a pressure gradient in the horizontal part to provide the centripetal acceleration:

$$\rho_{\rm air}\omega^2 r = \frac{dP}{dr}$$

• Shouldn't the water surface in the tank be non-flat, and the depth at D be below the average (rather than above)? Commented Aug 23, 2021 at 12:06
• @ChetMiller Maybe only the tube is rotating and not the entire tank.
– Tofi
Commented Aug 23, 2021 at 12:10
• @Tofi That wouldn't produce much of an effect. Commented Aug 23, 2021 at 12:17
• @ChetMiller It would still produce a pressure gradient in the tube because the air inside it would be rotating as well. I think, at equilibrium, the pressure at point A would be equal to atmospheric pressure, and the pressures at points B and C are related to that at A by the equation given in the answer.
– Tofi
Commented Aug 23, 2021 at 12:26
• or maybe pressure at point A is related to atmospheric pressure by Bernoulli's equation? idk.
– Tofi
Commented Aug 23, 2021 at 12:29